Quantum entanglement from a statistician’s perspective
Quantum entanglement is an intimidating phrase to encounter when you barely know what quantum means (and maybe it is even if you do). My daughter’s book, Quantum Entanglement for Babies, also does a good job of keeping the mystery alive:
Now I’ve just barely scratched the surface in quantum computing (and I mean barely, like I’ve gotten so far as to understand how to build a circuit to add two bits together. Yes, 1 + 1 = 2). But as I was going through the section on quantum entanglement in this tutorial, I immediately noticed something familiar that it was getting at (albeit in an unfamiliar, roundabout way). And that was statistical independence.
Some background
We can represent the state of qubits (like a bit, but in quantum), at a given point in time, as state vectors, which (loosely) correspond to the probability they will be measured in a particular state.
For example, suppose we have a qubit, \(q_0\), that has the following state vector:
\[q_0 = |0\rangle = \left[\begin{array}{c} 1 \\ 0 \\ \end{array}\right]\]
The positions of the vector represent the possible states the qubit can be in. Namely, since it’s basically just a bit, 0 (position 1) or 1 (position 2). The entries in the vector represent (again, loosely) the probability that the qubit will take on that state when measured. So in this example,
\[P(q_0 = 0) = 1 \hskip.1in P(q_0=1)=0\] It will always be measured in the 0 state.
Now suppose we introduce another qubit, \(q_1\). And remember, computers just store information as sequences of bits. This qubit can also only be measured in states 0 or 1. Thus, the possible bit strings are:
| \(q_1q_0\) | Represents the number… |
|---|---|
| 00 | \((0\times2^1) + (0 \times 2^0) = 0\) |
| 01 | \((0\times2^1) + (1 \times 2^0) = 1\) |
| 10 | \((1\times2^1) + (0 \times 2^0) = 2\) |
| 11 | \((1\times2^1) + (1 \times 2^0) = 3\) |
So one possible two-qubit state vector is:
\[|01\rangle = \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array}\right]\]
where, again, the positions represent the possible sequences of qubits (00, 01, 10, 11; there will always be \(2^n\) possible states, where \(n\) is the number of qubits), and the entries (for the third time, loosely) represent the probability of measuring that sequence. In this case,
\[P(q_0 = 1 \cap q_1 = 0) = 1; \hskip.1in P(\text{other combos}) = 0\] So now we can imagine the more interesting case where more than one entry is non-zero, that is, multiple different states have a positive probability of being measured. Given it still has to lead to a valid probability distribution, this means that the 100% must be distributed amongst the possibilities.
The final thing I’ll leave here is that the entries actually represent the square root of the probability, which is why I’ve been emphasizing probability “loosely”. So the “valid probability distribution” constraint applies to the square of the vector entries. In the first example above, a more complete way to write this would be:
\[P(q_0 = 0) = 1^2 = 1 \hskip.1in P(q_0=1) = 0^2 = 0\]
What is entanglement?
The tutorial has us consider a couple of two-qubit state vectors:
\[|\Phi^+\rangle = \frac{1}{\sqrt{2}} \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \\ \end{array}\right] \hskip.2in |+0\rangle = \frac{1}{\sqrt{2}} \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \\ \end{array}\right]\]
If we let \(X = q_1q_0\), that is, the bit string measured from the qubits, these imply the following:
\[P_{|\Phi^+\rangle}(X = 00) = P_{|\Phi^+\rangle}(X = 11) = \frac{1}{2}\]
\[P_{|+0\rangle}(X = 00) = P_{|+0\rangle}(X = 10) = \frac{1}{2}\]
Notice how both bits change in \(|\Phi^+\rangle\), but only one changes in \(|+0\rangle\). The former is entangled, the latter is not. This is because we cannot separate \(|\Phi^+\rangle\) into superpositions of two individual, one-qubit state vectors. But in \(|+0\rangle\), we can:
\[q_0 = \left[\begin{array}{c} 1 \\ 0 \\ \end{array}\right] = |0\rangle\] \[q_1 = \frac{1}{\sqrt{2}} \left[\begin{array}{c} 1 \\ 1 \\ \end{array}\right] = |+\rangle\]
Implying that \(q_0\) will always be measured to 0, and all uncertainty (random variability) lies in measuring \(q_1\). This is known as a product state, because the probabilities in the two-qubit state vector can be determined by a cross-product of the individual ones.
It’s just independence
Statistical independence occurs when the probability of observing an event does not change once we know something about another one. In our case, we can pretty clearly see this holds for \(|+0\rangle\) but not \(|\Phi^+\rangle\). Let’s look at the latter case.
From the the two-qubit state vector, we know the possible measurements are 00 or 11. Thus,
\[P_{|\Phi^+\rangle}(q_0 = 0) = P_{|\Phi^+\rangle}(q_0 = 1) = \frac{1}{2}\] \[P_{|\Phi^+\rangle}(q_1 = 0) = P_{|\Phi^+\rangle}(q_1 = 1) = \frac{1}{2}\]
Marginally, each qubit has an equal chance of being measured 0 or 1. But once we know something about the state of the other qubit, this changes:
\[P_{|\Phi^+\rangle}(q_1 = 0|q_0 = 0) = 1\] \[P_{|\Phi^+\rangle}(q_1 = 0|q_0 = 1) = 0\] \[P_{|\Phi^+\rangle}(q_1 = 1|q_0 = 0) = 0\] \[P_{|\Phi^+\rangle}(q_1 = 1|q_0 = 1) = 1\]
We could flip those around and condition \(q_0\) on \(q_1\) and we’d end up with the same result. What this shows is that in the entangled state,
\[P(q_0|q_1) \neq P(q_0)\] implying
\[P(q_0 \cap q_1) \neq P(q_0)P(q_1)\]
and therefore are not independent. Once we know (measure) one qubit, we automatically know what the other one will be. If you go through the same math for \(|+0\rangle\), you’ll see the marginal and conditional probabilities are in fact equal, and thus independent.
Now I don’t know if/how this might change once you start introducing more qubits or allow for the full range of phase, but to keep things intuitive, my working definition of quantum entanglement is:
Does the probability of a qubit being measured to a particular state depend on the state of another qubit? If yes, they are entangled; otherwise, they are not.